That is right, PF3 is polar, yet PF5 is non-polar.
To understand this, you need to know the atomic and molecular orbitals of phosphor. Briefly, in the valence shell of P (element of Group 15), there are 5 electrons. With 3 electrons from 3 bonding fluorine atoms, there are a total 8 electrons. The molecular orbits are SP3 hybridized thus molecular configuration has a shape of triangular pyramid. Notice that it is not tetrahedral, since the lone pair electrons is different from the P-F bonding. On the other hand, in PF5 molecule there are 10 electrons in the valence shell of P. The molecular orbits are sp3d hybridized thus molecular configuration has a shape of triangular bipyramid, which is symmetric thus the molecule is not polar.
2007-10-14 04:06:08
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answer #1
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answered by Hahaha 7
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VSEPR # lone pairs = ½[5(valence e- on P) - 3 (three P:F σ bonds)] = 1 lp It is an AX3E system trigonal pyramidal (ignore lp when naming) Put vectors along the P-F → δ- bonds but more importantly along the P-: → δ- unit. Do the vectors cancel; of course not and therefore the molecule is polar (Wikipedia has it at 1.03 D) but surprisingly the dominant → δ- is in the direction of the lone pair (I hope!). PF3 is a strange molecule probably because the Fs deactivate the lone pair. It is not a very reactive gas (only a slow rxn with water or O2 in air) and has ligating properties like CO. ;)
2016-05-22 08:14:46
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answer #3
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answered by ? 3
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